Question: Let $f(x)=x\sqrt{4-x}$, for $x<3$. Where does $f$ have critical points? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=0$ (Choice B) B $x=\dfrac{8}{3}$ (Choice C) C $x=\dfrac{16}{3}$ (Choice D) D $f$ has no critical points.
Answer: A critical point of $f$ is a point in the domain of $f$ where the derivative is either equal to zero or undefined. So in order to find the critical points of $f$, let's find its derivative. $\begin{aligned} f'(x)&=\dfrac{d}{dx}\left[ x\sqrt{4-x} \right] \\\\ &=x\cdot \dfrac{d}{dx}[\sqrt{4-x}]+ \dfrac{d}{dx}[x] \cdot \sqrt{4-x} \\\\ &=x\cdot \dfrac{1}{2\sqrt{4-x}}\cdot \dfrac{d}{dx}[4-x]+1\cdot \sqrt{4-x} \\\\ &=\dfrac{-x}{2\sqrt{4-x}}+\sqrt{4-x} \\\\ &=\dfrac{8-3x}{2\sqrt{4-x}} \end{aligned}$ Now let's look for $x$ -values where $f'$ is zero or undefined. $\dfrac{8-3x}{2\sqrt{4-x}}=0$ at $x=\dfrac{8}{3}$. $\dfrac{8-3x}{2\sqrt{4-x}}$ is undefined at $x=4$, but this is outside of our given domain. In conclusion, this is the only $x$ -value where $f$ has a critical point: $x=\dfrac{8}{3}$